Question

In a survey, 25 people were asked how much they spent on their child's last birthday gift. The results were roughly bell-shaped with a mean of $31 and standard deviation of $5. Construct a confidence interval at a 99% confidence level.

Give your answers to one decimal place.

Answer 1

To construct a confidence interval for the mean expenditure on a child's last birthday gift, we can use the formula for the confidence interval for the mean when the sample size is relatively small (n < 30) or when the population standard deviation is not known. However, since the sample size is 25 and the data is roughly bell-shaped, we can use the t-distribution for this calculation.

### Given:
- Sample size (n) = 25
- Mean (\(\bar{x}\)) = $31
- Standard deviation (s) = $5
- Confidence level = 99%

### Step 1: Determine the t-value

For a 99% confidence level with n - 1 degrees of freedom (df = 25 - 1 = 24), we look up the t-value in the t-distribution table or use a calculator.

For df = 24 at a 99% confidence level, the t-value is approximately **2.797**.

### Step 2: Calculate the standard error (SE)

\[
SE = \frac{s}{\sqrt{n}} = \frac{5}{\sqrt{25}} = \frac{5}{5} = 1
\]

### Step 3: Calculate the margin of error (ME)

\[
ME = t \times SE = 2.797 \times 1 = 2.797
\]

### Step 4: Construct the confidence interval

\[
\text{Lower limit} = \bar{x} - ME = 31 - 2.797 \approx 28.2
\]
\[
\text{Upper limit} = \bar{x} + ME = 31 + 2.797 \approx 33.8
\]

### Final Result

The 99% confidence interval for the mean amount spent on a child's last birthday gift is approximately:

\[
\boxed{(28.2, 33.8)}
\]